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Ann [662]
3 years ago
7

Cost 75.09 with a 38% markup

Mathematics
1 answer:
Sophie [7]3 years ago
5 0

Answer:

103.62

Step-by-step explanation:

If I am understanding this question correctly you want the answer for how much the total would be for a 38% markup at a starter price of 75.09.

In which case it would be 103.62.

It is a 28.53 increase.

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A savings account earns 4.5% simple interest per year. If $650 is deposited and no withdrawals are made during one year, how muc
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679.25 is the answer
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You decide to take your car on a drive through Canada, where gas is sold in liters and distances are measured in kilometers. Sup
12345 [234]

Answer:

the no of liters required is 50.55 liters

Step-by-step explanation:

Given that

The car gas efficiency is 23.0 mi/gal

distance is 495 km

1 gal = 3.78 L

1 km = 0.6214 mi

We need to find out the no of liters of gas that would needed to complete a trip

So,

= (495 × 3.78 × 0.6214) ÷ (23)

= 50.55 Liters

hence, the no of liters required is 50.55 liters

7 0
3 years ago
Fill in the table using this function rule
tigry1 [53]
This one isn't too hard, just plug in each value!

x = -1
y = -4(-1) + 3
y = 7

x = 0
y = -4(0) + 3
y = 3

x = 1
y = -4(1) + 3
y = -1

x = 2
y = -4(2) + 3
y = -5
8 0
3 years ago
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QUESTION 3 [10 MARKS] A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the nu
Andru [333]

Answer:

(a)Revenue=580x-10x²,

Marginal Revenue=580-20x

(b)Fixed Cost, =900

Marginal Cost,=300+50x

(c)Profit Function=280x-900-35x²

(d)x=4

Step-by-step explanation:

The price, p = 580 − 10x where x is the number of cakes sold per day.

Total cost function,c = (30+5x)²

(a) Revenue Function

R(x)=x*p(x)=x(580 − 10x)

R(x)=580x-10x²

Marginal Revenue Function

This is the derivative of the revenue function.

If R(x)=580x-10x²,

R'(x)=580-20x

(b)Total cost function,c = (30+5x)²

c=(30+5x)(30+5x)

=900+300x+25x²

Therefore, Fixed Cost, =900

Marginal Cost Function

This is the derivative of the cost function.

If c(x)=900+300x+25x²

Marginal Cost, c'(x)=300+50x

(c)Profit Function

Profit, P(x)=R(x)-C(x)

=(580x-10x²)-(900+300x+25x²)

=580x-10x²-900-300x-25x²

P(x)=280x-900-35x²

(d)To maximize profit, we take the derivative of P(x) in (c) above and solve for its critical point.

Since P(x)=280x-900-35x²

P'(x)=280-70x

Equate the derivative to zero

280-70x=0

280=70x

x=4

The number of cakes that maximizes profit is 4.

5 0
3 years ago
HELPPPPPPPPPPPP
pochemuha
<span>14% per year
</span>14% /12 = 1.2%  is <span> the monthly growth rate</span>
3 0
3 years ago
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